Permutations and Combinations

Rosen 4.3

Permutations

• A permutation of a set of distinct objects is an ordered arrangement these objects.

• An ordered arrangement of r elements of a set is called an r-permutation.

• The number of r-permutations of a set with n elements is denoted by P(n,r).

A = {1,2,3,4} 2-permutations of A include 1,2; 2,1; 1,3; 2,3; etc…

Counting Permutations

• Using the product rule we can find P(n,r)

= n*(n-1)*(n-2)* …*(n-r+1)

= n!/(n-r)!

How many 2-permutations are there for the set {1,2,3,4}? P(4,2)

12!2

!4

1*2

1*2*3*43*4

Combinations

• An r-combination of elements of a set is an unordered selection of r element from the set. (i.e., an r-combination is simply a subset of the set with r elements).

Let A={1,2,3,4} 3-combinations of A are{1,2,3}, {1,2,4}, {1,3,4}, {2,3,4}(same as {3,2,4})• The number of r-combinations of a set with n

distinct elements is denoted by C(n,r).

ExampleLet A = {1,2,3}

2-permutations of A are: 1,2 2,1 1,3 3,1 2,3 3,2

6 total. Order is important

2-combinations of A are: {1,2}, {1,3}, {2,3}

3 total. Order is not important

If we counted the number of permutations of each 2-combination we could figure out P(3,2)!

How to compute C(n,r)• To find P(n,r), we could first find C(n,r),

then order each subset of r elements to count the number of different orderings. P(n,r) = C(n,r)P(r,r).

• So C(n,r) = P(n,r) / P(r,r)

)!(!

!

!)!(

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rnr

n

rrn

rrn

rrr

rnn

A club has 25 members.

• How many ways are there to choose four members of the club to serve on an executive committee?– Order not important

– C(25,4) = 25!/21!4! = 25*24*23*22/4*3*2*1 =25*23*22 = 12,650

• How many ways are there to choose a president, vice president, secretary, and treasurer of the club?– Order is important

– P(25,4) = 25!/21! = 303,600

The English alphabet contains 21 consonants and 5 vowels. How many strings of six lower case

letters of the English alphabet contain:

• exactly one vowel?

• exactly 2 vowels

• at least 1 vowel

• at least 2 vowels

The English alphabet contains 21 consonants and 5 vowels. How many strings of six lower case

letters of the English alphabet contain:

• exactly one vowel?Note that strings can have repeated letters!We need to choose the position for the vowel

C(6,1) = 6!/1!5! This can be done 6 ways.Choose which vowel to use. This can be done in 5 ways.Each of the other 5 positions can contain any of the 21

consonants (not distinct).There are 215 ways to fill the rest of the string.

6*5*215

The English alphabet contains 21 consonants and 5 vowels. How many strings of six lower case

letters of the English alphabet contain:

• exactly 2 vowels?Choose position for the vowels.

C(6,2) = 6!/2!4! = 15

Choose the two vowels.

5 choices for each of 2 positions = 52

Each of the other 4 positions can contain any of 21 consonants.

214

15*52*214

The English alphabet contains 21 consonants and 5 vowels. How many strings of six lower case

letters of the English alphabet contain:

• at least 1 vowel

Count the number of strings with no vowels and subtract this from the total number of strings.

266 - 216

The English alphabet contains 21 consonants and 5 vowels. How many strings of six lower case

letters of the English alphabet contain:

• at least 2 vowels

Compute total number of strings and subtract number of strings with no vowels and the number of strings with exactly 1 vowel.

266 - 216 - 6*5*215

Corollary 1: Let n and r be nonnegative integers with r n. Then C(n,r) = C(n,n-r)

Proof:

C(n,r) = n!/r!(n-r)!

C(n,n-r) = n!/(n-r)!(n-(n-r))! = n!/r!(n-r)!

Binomial Coefficient

Another notation for C(n,r) is . This number is also called a binomial coefficient.

These numbers occur as coefficients in the expansions of powers of binomial expressions such as (a+b)n.

n

r

Binomial Theorem

Let x and y be variables and let n be a positive integer. Then

(x y)n C(n, j)xn j

j0

n

y j

n

0

xn

n

1

xn 1y

n

2

xn 2y2 ...

n

n 1

xyn1

n

n

yn

Pascal’s IdentityLet n and k be positive integers with n k.

Then C(n+1,k) = C(n, k-1) + C(n,k).

Proof:

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)1(!

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)!)(1()!1(

!

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!

),()1,(

knC

knk

n

knk

nn

knk

knkn

knknk

nkn

knknkk

kn

knk

n

knk

n

knCknC

Let n be a positive integer. Then

Proof: We know from set theory that the number of subsets in a set of size n is 2n. We also know that C(n,k) is the number of subsets of a set of size n that are of size k.

counts the number of subsets

of every size from 0 (empty set) to n. Therefore the sum must add up to 2n.

C(n,k) 2n

k0

n

C(n,k)

k0

n

Vandermonde’s Identity

C(mn, r) C(m, r k)

k0

r

C(n,k).

Proof: Suppose there are n items in one set and m items in a second set. (The sets are distinct). Then the total number of ways to pick r elements from the union of these sets is C(m+n,r).

Another way to pick r elements from the union is to pick k elements from the first set and then r-k elements from the second set, where 0 k r. There are C(n,k) ways to pick the k elements from the first set and C(m,r-k) ways to pick the rest of the elements from the second set.

Proof: Suppose there are n items in one set and m items in a second set. Then the total number of ways to pick r elements from the union of these sets is C(m+n,r).

Another way to pick r elements from the union is to pick k elements from the first set and then r-k elements from the second set, where 0 k r. For any k,there are C(n,k) ways to pick the k elements from the first set and C(m,r-k) ways to pick the rest of the elements from the second set. By the product rule there are C(m,r-k)C(n,k) ways to pick r elements for a particular k. For all possible values of k

C(mn, r) C(m, r k)

k0

r

C(n,k).

C(m, r k)

k0

r

C(n, k).

Pascal’s Triangle0

0

1

0

1

1

2

0

2

1

2

2

3

0

3

1

3

2

3

3

1

1 1

1

1 2

3 3

1

1

1 4 6 4 1

n’th row, Cnk =k = 0, 1, …, n

n

r